Question

已知三点A[2,3]B[-1,-1]C[6,k],其中k为常数,若|AB|=|AC|.则向量AB与向量AC的夹角是?

Answer 1

|AB|=5,

|AC|^2=16+(k-3)^2=25,

(k-3)^2=9,k1=0,k2=6.

AB=(-3,-4)

k1=0时AC=(4,-3),

向量AB*AC=0,∠BAC=90°；

k2=6时AC=(4,3),AB*AC=-24,

cosBAC=-24/25,

∠BAC=180°-arccos(24/25).